Question

Helium nucleus is moving in a circular path of radius 0.8m. If it takes 2 sec to complete one revolution. Find out magnetic field produced at the centre of the circle.

Answer 1

application of magnetic field at center of current carrying circular coil

Answer 2

Answer:

The magnetic field at the center if the circle is $B=1.25\times 10^{-25}\ T$

Explanation:

It is given that,

Radius of the circular path, r = 0.8 m

Time taken to complete one revolution, t = 2 s

We need to find the magnetic field produced at the center of the circle. The formula for magnetic field at the center of the circle is given by :

$B=\dfrac{\mu_oI}{2r}$

Since, $I=\dfrac{q}{t}$

For helium nucleus, q = 2e

$B=\dfrac{\mu_o\times 2e}{2rt}$

$B=\dfrac{4\pi \times 10^{-7}\times 2\times 1.6\times 10^{-19}}{2\times 0.8\ m\times 2\ s}$

$B=1.25\times 10^{-25}\ T$

So, the magnetic field produced at the center of the circle is $B=1.25\times 10^{-25}\ T$. Hence, this is the required solution.